这是我第一次尝试使用mySQL,并且在尝试将我的jotform php链接到mySQL数据库时遇到了一个问题。
这是我对数据库的查询:
$query = "SELECT * FROM `new` WHERE `submission_id` = '$submission_id'";
$sqlsearch =mysql_query ($query) or die(mysql_error());
$resultcount = mysql_numrows($sqlsearch);
if ($resultcount > 0) {
mysql_query("UPDATE `new` SET
`captainsfull` = '$captainsfull',
`email14` = '$email14',
`teamcaptains` = '$teamcaptains',
`teammate2` = '$teammate2',
`teammate3` = '$teammate3',
`teammate4` = '$teammate4',
`teammate511` = '$teammate511',
WHERE `submission_id` = '$submission_id'")
or die(mysql_error());
} else {
mysql_query("INSERT INTO `new` (submission_id, formID, IP,
captainsfull, email14, teamcaptains, teammate2, teammate3, teammate4, teammate511)
VALUES ('$submission_id', '$formID', '$ip',
'$captainsfull', '$email14', '$teamcaptains', '$teammate2', '$teammate3', '$teammate4', '$teammate511') ")
or die(mysql_error());
}
第4行错误消息“near')'中的线索”指向else块中INSERT语句中的问题。
我没有看到任何明显的错误。输出查询字符串以及mysql错误对于调试非常有用。就像是:
$qry = "INSERT INTO `new`
(submission_id, formID, IP, captainsfull, email14, teamcaptains,
teammate2, teammate3, teammate4, teammate511)
VALUES ('$submission_id', '$formID', '$ip', '$captainsfull', '$email14',
'$teamcaptains', '$teammate2', '$teammate3', '$teammate4', '$teammate511') ";
mysql_query($qry) or die($qry . mysql_error());
(如果可能的话,请考虑从已弃用的mysql函数迁移到mysqli或PDO。)